Question

An upright image which reduced in size 10 times occurred in a mirror. If the radius of

curvature of the mirror is 2 m, bow far is the object from the mirror?
How to solve?​

Answer 1

p = -9 m

Step-by-step explanation:

For this exercise we use the equation of the geometric optics constructor

`(1)/(f) = (1)/(p) + (1)/(q)`

where f is the focal length, p and q are the distance to the object and the image, respectively.

The mirrors the focal length is

f = R / 2

f = 2/2

f = 1 m

the magnification is

m =
`(h')/(h) = - (q)/(p)`

indicates that the image was reduced h ’= h/10 implies that m = 1/10

`(1)/(10) = - (q)/(p)`

we write our system of equations

p = -10q

1/1 =
`(1)/(p) + (1)/(q)`

we substitute

1 =
`(1)/(p) - (10)/(p)`

1 = 1/p (1 - 10)

1 = -9 / p

p = -9 m